Laws of Motion Test

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Laws of Motion Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

Banking of roads is provided at turns to :

A
reduce intertia

B
increase centripetal force

C
increase fiction

D
none of these

Solution

Banking a road changes the angle of the normal reaction force, this change creates a component of the normal reaction along the radius of curvature of the road towards the centre. This provides additional centripetal force.
Question 2
1 / -0

A large force is acting on a body for a short time. The impulse imparted is equal to the change in :

A
Acceleration

B
Momentum

C
Energy

D
Velocity

Solution

If a large force F acts for a short time dt the impulse imparted I is
$$I=Fdt=\dfrac{dp}{dt}dt$$
$$I=dp$$ = change in momentum
Question 3
1 / -0

A cyclist is moving on a circular track of radius $$80\ m$$ with a velocity of $$72\ km/hr$$. He has to lean from he vertical approximately through an angle-

A
$$\tan ^{ -1 }{ \left( 1/4 \right) } $$

B
$$\tan ^{ -1 }{ \left( 1 \right) } $$

C
$$\tan ^{ -1 }{ \left( 1/2 \right) } $$

D
$$\tan ^{ -1 }{ \left( 2 \right) } $$

Solution

It is given that,

$$ v=72Km/hr $$

$$ =\dfrac{5}{18}\times 72=20\,m/s $$

$$ radius\,\,r=80\,m $$

Bending of a cyclist in circular motion is given by the relation as
$$ \tan \theta =\dfrac{{{v}^{2}}}{rg} $$
$$ \tan \theta =\dfrac{20\times 20}{80\times 10} $$ $$ \theta ={{\tan }^{-}}(\dfrac{1}{2}) $$
Question 4
1 / -0

When the speed of a moving body is doubled:-

A
Its acceleration is doubled

B
Its momentum is doubled

C
Its kinetic energy is doubled

D
Its potential energy is doubled

Solution

Its momentum gets doubled.
momentum is mass times speed
$$p = mv$$
Therefore when speed is doubled, momentum $$‘p’$$gets doubled.
$$p’ = 2p$$
Question 5
1 / -0

What is the difference of force of friction acting on a body moving on a fixed surface?

A
Along the different of motion

B
Opposite to the direction of motion

C
Independent of Motion

D
Both (A) and (B)

Solution

When we try to slide a body on a surface, the motion of the body is opposed by a force called the force of friction. The frictional force arises due to intermolecular interaction.
When an external force $$(F)$$ is applied to move the body and the body does not move, then the frictional force acts opposite to applied force $$F$$ and is equal to the applied force i.e., $$F-f=0$$ frictional force ,$$f$$ and applied force $$F$$ . When the body remains at rest, the frictional force is called the static friction. Static friction is a self-adjusting force.
Friction force, $${{f}_{r}}=\mu mg$$
Where,
$$\mu $$ = coefficient of friction.

Hence, friction force act opposite to the direction of motion.
Question 6
1 / -0

A brick of mass $$m$$ , tied to a rope , is being whirled in a vertical circle, with a uniform speed. The tension in the rope is:

A
The same throughout

B
Largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point

C
Largest when the rope is horizontal and smaller when it is vertical.

D
Largest when the brick is at the lowest point of the circular path and smallest when it is at the highest point

Solution

$$ T_H + mg = m \omega^2 r$$
$$ T_L - mg = m\omega^2 r$$
$$ \therefore T_L > T_H$$
Question 7
1 / -0

When the constant force is 10 N for time period 1 s to 10 second then, find the impulse?

A
45 Ns

B
50 Ns

C
90 Ns

D
100 Ns

Solution

Constant force, $$F=10\,N$$
Time period, $${{t}_{2}}=10\,s\,\,and\,{{t}_{1}}=1s$$
Impulse, $$I=\int{F.dt}=F\int{dt}$$ Area of F and t graph.
Where,
$$F$$=force
$$t$$=time
$$I=F({{t}_{2}}-{{t}_{1}})=10(10-1)=90\,Ns$$
Hence, impulse is $$90\,Ns$$
Question 8
1 / -0

Force acting on a body is given by $$F=(1200-4\times 10^{5}t)\ N$$. After starting the motion to it moves with constant velocity, how much impulse of force is acting on it?

A
Zero

B
$$0.9\ Ns$$

C
$$1.8\ Ns$$

D
$$3.6\ Ns$$

Solution

zero.

a body moving with constant velocity has no net force acting on it.
Question 9
1 / -0

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If $$T_{1} and \ T_{2}$$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

A
$$T_{2}-T_{1}=6mg$$

B
$$T_{2}-T_{1}=4mg$$

C
$$T_{2}-T_{1}=2mg$$

D
$$T_{2}-T_{1}=mg$$

Solution

Centripetal Force is: $$\dfrac { m{ v }^{ 2 } }{ r } $$
At the highest point: $${ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } -mg$$
At the lowest point: $${ T }_{ 2 }=\dfrac { m{ v }^{ 2 } }{ r } +mg$$
$$\therefore \quad { T }_{ 2 }-{ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } +mg-(\dfrac { m{ v }^{ 2 } }{ r } -mg)=2mg$$
Question 10
1 / -0

Statement I: The max value of force F such that the block shown does not move is $$\dfrac{\mu mg}{cos\theta }$$ where $$\mu $$ is the coefficient of friction between block end surface.
Statement II: Frictional Force = (Coefficient of friction) x (Normal reaction)

A
Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1

B
Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1

C
Statement 1 is True, Statement 2 is False

D
Statement 1 is False, Statement 2 is True

Solution

The max value of force F such that the block shown does not move is:
$$Fcos\theta= \mu(mg+Fsin \theta)$$
$$F= \dfrac{\mu m g}{(cos\theta -\mu sin \theta)}$$
So, the statement 1 is false.
The frictional force is proportional to the normal reaction.
So, Frictional Force = (Coefficient of friction) x (Normal reaction)

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 27

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Laws of Motion Test - 27

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SHARING IS CARING

Question 1

reduce intertia

increase centripetal force

increase fiction

none of these

Solution

Question 2

Acceleration

Momentum

Energy

Velocity

Solution

Question 3

$$\tan ^{ -1 }{ \left( 1/4 \right) } $$

$$\tan ^{ -1 }{ \left( 1 \right) } $$

$$\tan ^{ -1 }{ \left( 1/2 \right) } $$

$$\tan ^{ -1 }{ \left( 2 \right) } $$

Solution

Question 4

Its acceleration is doubled

Its momentum is doubled

Its kinetic energy is doubled

Its potential energy is doubled

Solution

Question 5

Along the different of motion

Opposite to the direction of motion

Independent of Motion

Both (A) and (B)

Solution

Question 6

The same throughout

Largest when the brick is at the highest point of the circular path and smallest when it is at the lowest point

Largest when the rope is horizontal and smaller when it is vertical.

Largest when the brick is at the lowest point of the circular path and smallest when it is at the highest point

Solution

Question 7

45 Ns

50 Ns

90 Ns

100 Ns

Solution

Question 8

Zero

$$0.9\ Ns$$

$$1.8\ Ns$$

$$3.6\ Ns$$

Solution

Question 9

$$T_{2}-T_{1}=6mg$$

$$T_{2}-T_{1}=4mg$$

$$T_{2}-T_{1}=2mg$$

$$T_{2}-T_{1}=mg$$

Solution

Question 10

Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1

Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1

Statement 1 is True, Statement 2 is False

Statement 1 is False, Statement 2 is True

Solution

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